Question

Математики, влетайте =D

Вычислить предел, пользуясь правилом Лопиталя:

Answer 1

$displaystyle lim_{x to 0}dfrac{6sin 2x-12x}{x^3}=lim_{x to 0}dfrac{(6sin 2x-12x)'_x}{(x^3)'_x}=lim_{x to 0}frac{6cdot 2cos 2x-12}{3x^2}=\ \ \ =lim_{x to 0}frac{(12cos 2x-12)'}{(3x^2)'}=lim_{x to 0}frac{-24sin2x}{6x}=lim_{x to 0}frac{(-24sin 2x)'}{(6x)'}=\ \ \ =lim_{x to 0}frac{-48cos 2x}{6}=frac{-48}{6}=-8$